Remove Letter To Equalize Frequency — LeetCode 2423 Python Solution

The problem

You are given a 0-indexed string word, consisting of lowercase English letters. You need to select one index and remove the letter at that index from word so that the frequency of every letter present in word is equal.

Example

Input

word = "abcc"

Output

true

Explanation

Select index 3 and delete it: word becomes "abc" and each character has a frequency of 1.

Python solution

class Solution:
    def equalFrequency(self, word: str) -> bool:
        cnt = Counter(word)
        for c in cnt.keys():
            cnt[c] -= 1
            if len(set(v for v in cnt.values() if v)) == 1:
                return True
            cnt[c] += 1
        return False

Complexity

Pattern: Hash Map

Trade memory for time: remember what you have seen so the second pass never happens. LeetCode 2423. Remove Letter To Equalize Frequency is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Hash Table and Counting.

The hash map guide has the Python template for the pattern and the 709 LeetCode problems that use it.

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